Precalculus, Chapter 4, 4.7, Section 4.7, Problem 51

By definition, arcsin x = y if sin y = x . Replacing x by sin y, yields:
arcsin x = arcsin (sin y) = y
Hence, replacing y by 3pi , yields:
arcsin(sin 3pi) = 3pi
You need to evaluate arcsin(sin 3pi) , hence, you need to verify if sin 3pi falls in interval [-1,1].
Hence, first you need to evaluate sin 3pi, such that:
sin 3pi = sin(2pi+pi) = sin 2pi*cos pi + sin pi*cos 2pi = 0*(-1) + 0*1 = 0 Notice that 0 in [-1,1].
Replacing sin 3pi by 0, yields:
arcsin(sin 3pi) = arcsin 0 = 0, pi, 2pi, 3pi,...,n*pi,..
Hence, arcsin 0 = k*pi, where k in Z , so arcsin (sin 3pi) = 3pi .